the odd Bernoulli numbers are zero

Recall that, for $k\ge 0$, the Bernoulli numbers $B_k$ are defined as the coefficients in the Taylor expansion:

${\displaystyle \frac{t}{e^t-1}}={\displaystyle \sum _{k\ge 0}}{B}_k{\displaystyle \frac{t^k}{k!}}.$(1)

Just to name a few:

$$B_0=1,B_1=-\frac{1}{2},B_2=\frac{1}{6},B_3=0,B_4=-\frac{1}{30},B_5=0,\mathrm{\dots },B_{10}=\frac{5}{66},\mathrm{\dots }$$

Lemma.

If $k\mathrm{\ge }\mathrm{3}$ is odd then $B_k\mathrm{=}\mathrm{0}$.

Proof.

From the right hand side of (1) we extract the term corresponding to $k=1$:

${\displaystyle \frac{t}{e^t-1}}=-{\displaystyle \frac{t}{2}}+{\displaystyle \sum _{k\ge 0,k\ne 1}}{B}_k{\displaystyle \frac{t^k}{k!}}.$(2)

Thus:

${\displaystyle \frac{t}{e^t-1}}+{\displaystyle \frac{t}{2}}={\displaystyle \sum _{k\ge 0,k\ne 1}}{B}_k{\displaystyle \frac{t^k}{k!}}$(3)

and the left hand side can be rewritten as:

${\displaystyle \frac{t}{e^t-1}}+{\displaystyle \frac{t}{2}}={\displaystyle \frac{2t+t(e^t-1)}{2(e^t-1)}}={\displaystyle \frac{t}{2}}\cdot {\displaystyle \frac{e^t+1}{e^t-1}}={\displaystyle \frac{t}{2}}\cdot {\displaystyle \frac{e^{t/2}+e^{-t/2}}{e^{t/2}-e^{-t/2}}}.$(4)

Hence, if one replaces $t$ by $-t$ then (4) is unchanged. Since (4) is the left hand side of (3), the quantity

$$\sum _{k\ge 0,k\ne 1}{B}_k\frac{t^k}{k!}$$

is also unchanged when $t$ is exchanged by $-t$, and so we must have $B_k={(-1)}^k{B}_k$ for $k\ne 1$. We conclude that if $k\ge 3$ and $k$ is odd, $B_k=0$.∎